Problem: Let $h(x)=\dfrac{\csc(3x)}{x^3}$. Find $h'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-3\csc(3x)\Bigl(x\cot(3x)+1\Bigr)}{x^4}$ (Choice B) B $\dfrac{-\csc(3x)\Bigl(x\cot(3x)+3\Bigr)}{x^6}$ (Choice C) C $\dfrac{-\csc(3x)\cot(3x)}{3x^2}$ (Choice D) D $\dfrac{-\csc(3x)\cot(3x)}{x^2}$
Explanation: $h$ is a quotient of a composite function and another function. Let... $u(x)=\csc(x)$ $v(x)=3x$ $w(x)=x^3$... then $h(x)=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $h'(x)$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=-\csc(x)\cot(x)$ $v'(x)=3$ $w'(x)=3x^2$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{-\csc(3x)\cot(3x)}\cdot3\cdot x^3-{\csc(3x)}(3x^2)}{(x^3)^2} \\\\ &=\dfrac{-3x^3\csc(3x)\cot(3x)-3x^2\csc(3x)}{x^6} \\\\ &=\dfrac{-3x^2\csc(3x)\Bigl(x\cot(3x)+1\Bigr)}{x^6} \gray{\text{Simplify}} \\\\ &=\dfrac{-3\csc(3x)\Bigl(x\cot(3x)+1\Bigr)}{x^4} \gray{\text{Cancel common factors}} \end{aligned}$ In conclusion, $h'(x)=\dfrac{-3\csc(3x)\Bigl(x\cot(3x)+1\Bigr)}{x^4}$